`|x+1|=|x-3|`
`⇒`\(\left[ \begin{array}{l}x+1=x-3\\x+1=-(x-3)\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x+1=x-3\\x+1=-x+3\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x-x=-3-1\\x+x=3-1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}0x=-4\text{(vô lý)}\\x+x=3-1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}0x=-4\text{(vô lý)}\\2x=2\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}0x=-4\text{(vô lý)}\\x=1\end{array} \right.\)
Vậy `x∈{1}.`